I have a function of the form $f(x)=\sum^x_{k=1}\lfloor \frac{1}{1+(x \bmod k)} \rfloor$ and I want to know whether it's possible to simplify this. I think that since the $\bmod$ operator can be rephrased as a function involving a floor function that maybe these can cancel out?
2026-04-03 14:51:07.1775227867
Can you simplify a mod operator in the denominator of a floored fraction?
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